A ball (initially at rest) falls vertically for 2 s and hits a smooth plane inclined at 30 ° to the…

1. $40.63 \mathrm{m}$
2. $20.63 \mathrm{m}$
3. $30.63 \mathrm{m}$
4. $50.63 \mathrm{m}$

Solution

Let the ball strike the plane at A. We assign a co-ordinate system with origin at A. The velocity of the ball while striking the 30^o plane = gt = 2g m/s. The components of u along and perpendicular to the plane are u sin 30^o and u cos 30^o respectively. After impact, the velocity component u sin 30^o along the plane remains unaffected while the component u cos 30^o perpendicular to the plane becomes e(u cos 30^o) on rebound.

Thus velocity of rebound$=e\text{.}2\text{g}\text{.}\frac{\sqrt{3}}{2}=\sqrt{3}\text{eg}$



The acceleration perpendicular to the plane$=-\text{g}\text{cos}30^{\circ }=-\text{g}\sqrt{3}/2$

If t be the time of striking the plane again, then

$0=\sqrt{3}\mathit{\text{egt}}-\frac{1}{2}\frac{\sqrt{3}}{2}\text{g}{t}^2$

⇒ $t=4e=4\times \frac{5}{8}=2\text{.}5\text{sec}$

The distance $\text{AB}=u\text{sin}30^o\mathrm{ t}+\frac{1}{2}\text{g}\text{sin}30^o{t}^2$

$=\left(2\text{g}\right)\times 2\text{.}5\times \frac{1}{2}+\frac{1}{2}\times 10\times \frac{1}{2}{\left(\frac{5}{2}\right)}^2$

$=25+\frac{250}{16}=40\text{.}63\text{m}$

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